TITLE: Grothendieck group for projective space over the dual numbers
QUESTION [10 upvotes]: Fix a field $k$.  For a singular variety $X$, I understand that the Grothendieck group $K^0(X)$ of vector bundles on $X$ is not necessarily isomorphic to the Grothendieck group $K_0(X)$ of coherent sheaves on $X$.  
I am curious to learn what is known about these two groups in one family of examples: $\mathbb P^n_{D}$, where $D$ is the dual numbers $D=k[\epsilon]/(\epsilon^2)$.  
References would be especially appreciated, as I know very little about K-theory.

REPLY [9 votes]: If $X$ is a noetherian separated scheme and $X_{red}$ its reduction , we have $K_0(X)=K_o(X_{red})$: in other words  $K_o$ doesn't see  nilpotents .
Much more generally and profoundly, Quillen has proved that for all his $K$-theory groups, $K_i(X)=K_i(X_{red})$.
In your particular case you thus have (in the following $T$ is an indeterminate)
$$K_0(\mathbb P^n_D)=K_0(\mathbb P^n_k)=\mathbb Z[T]/(T^{n+1})$$
As for $K^0$, a special case of a theorem of Berthelot (SGA 6, Exposé VI, Théorème 1.1, page 365)
states that, for any commutative ring $A$, we have  $K^0(\mathbb P^n_A)=K^o(A)[T]/(T^{n+1})$.
If $A=D=k[\epsilon]$, we have $K^0(D)=\mathbb Z$, since projective modules over local rings (like $D$) are free.
So here too $$K^0(\mathbb P^n_D)=\mathbb Z[T]/(T^{n+1})$$
Bibliography
Srinivas has written this  nice book on $K$-theory.  
And as an homage to the recently sadly departed Daniel Quillen, let me refer to his groundbreaking paper
  "Higher algebraic $K$-theory I",  published in Springer's Lecture Notes LNM 341.